Proof: Get Begin Expr Params 26

Let's prove the following theorem:

if the following are true:

expression state at time 26 = "begin_expr"
the expression at time 26 = self."y" = 0
arguments stack at time 26 = [ ]

then arguments stack at time 27 = [ [ 0, [ ] ], [ ] ]

Proof:

View as a tree | View dependent proofs | Try proving it

Given

1	expression state at time 26 = "begin_expr"
2	the expression at time 26 = self."y" = 0
3	arguments stack at time 26 = [ ]

Proof Table

#	Claim	Reason
1	arguments stack at time (26 + 1) = [ [ 0, [ ] ], arguments stack at time 26 ]	if expression state at time 26 = "begin_expr" and the expression at time 26 = self."y" = 0, then arguments stack at time (26 + 1) = [ [ 0, [ ] ], arguments stack at time 26 ] (Set Attribute Begin Arguments Property)
2	[ [ 0, [ ] ], arguments stack at time 26 ] = [ [ 0, [ ] ], [ ] ]	if arguments stack at time 26 = [ ], then [ [ 0, [ ] ], arguments stack at time 26 ] = [ [ 0, [ ] ], [ ] ] (Substitution)
3	arguments stack at time (26 + 1) = [ [ 0, [ ] ], [ ] ]	if arguments stack at time (26 + 1) = [ [ 0, [ ] ], arguments stack at time 26 ] and [ [ 0, [ ] ], arguments stack at time 26 ] = [ [ 0, [ ] ], [ ] ], then arguments stack at time (26 + 1) = [ [ 0, [ ] ], [ ] ] (Transitive Property of Equality)
4	26 + 1 = 27	26 + 1 = 27 (Fixed Value Statement)
5	arguments stack at time (26 + 1) = arguments stack at time 27	if 26 + 1 = 27, then arguments stack at time (26 + 1) = arguments stack at time 27 (Substitution)
6	arguments stack at time 27 = [ [ 0, [ ] ], [ ] ]	if arguments stack at time (26 + 1) = arguments stack at time 27 and arguments stack at time (26 + 1) = [ [ 0, [ ] ], [ ] ], then arguments stack at time 27 = [ [ 0, [ ] ], [ ] ] (Transitive Property of Equality Variation 2)

Comments

Please log in to add comments